Introduction to Open Data Science 2018 is a course about modern methods and tools in data science.
You can find my course github repository here https://github.com/eavalo/IODS-project/.
Read in the dataset. The dataset contains 7 variables:
learning2014 <- read.csv("~/git/IODS-project/data/learning2014.csv")
Check that the data was read correctly. Print out the first few rows and check the type of the columns:
head(learning2014)
## gender age attitude deep surf stra points
## 1 F 53 37 3.583333 2.583333 3.375 25
## 2 M 55 31 2.916667 3.166667 2.750 12
## 3 F 49 25 3.500000 2.250000 3.625 24
## 4 M 53 35 3.500000 2.250000 3.125 10
## 5 M 49 37 3.666667 2.833333 3.625 22
## 6 F 38 38 4.750000 2.416667 3.625 21
str(learning2014)
## 'data.frame': 166 obs. of 7 variables:
## $ gender : Factor w/ 2 levels "F","M": 1 2 1 2 2 1 2 1 2 1 ...
## $ age : int 53 55 49 53 49 38 50 37 37 42 ...
## $ attitude: int 37 31 25 35 37 38 35 29 38 21 ...
## $ deep : num 3.58 2.92 3.5 3.5 3.67 ...
## $ surf : num 2.58 3.17 2.25 2.25 2.83 ...
## $ stra : num 3.38 2.75 3.62 3.12 3.62 ...
## $ points : int 25 12 24 10 22 21 21 31 24 26 ...
Explore the data by plotting
library(GGally)
library(ggplot2)
# Define the plot.
p <- ggpairs(learning2014, mapping = aes(col = gender, alpha=0.3),
lower = list(combo = wrap("facethist", bins = 20)), legend=1)
# Draw the plot
p
There are almost twice as many women in the dataset compared to men. The distribution of age is skewed towards higer values. The distribution of attitutde, deep, surf, stra and points seems to be more normally distributed. The variable with the highest correlation with points is attitude.
summary(learning2014$gender)
## F M
## 110 56
summary(learning2014$age)
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 17.00 21.00 22.00 25.51 27.00 55.00
summary(learning2014$attitude)
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 14.00 26.00 32.00 31.43 37.00 50.00
summary(learning2014$deep)
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 1.583 3.333 3.667 3.680 4.083 4.917
summary(learning2014$surf)
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 1.583 2.417 2.833 2.787 3.167 4.333
summary(learning2014$stra)
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 1.250 2.625 3.188 3.121 3.625 5.000
summary(learning2014$points)
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 7.00 19.00 23.00 22.72 27.75 33.00
Fit a regression model using points as the outcome variable and age, gender and attitude as explanatory variables:
# Fit the linear regression model
lm_model <- lm(points ~ age + gender + attitude, data=learning2014)
# Sumamry of the model
summary(lm_model)
##
## Call:
## lm(formula = points ~ age + gender + attitude, data = learning2014)
##
## Residuals:
## Min 1Q Median 3Q Max
## -17.4590 -3.3221 0.2186 4.0247 10.4632
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 13.42910 2.29043 5.863 2.48e-08 ***
## age -0.07586 0.05367 -1.414 0.159
## genderM -0.33054 0.91934 -0.360 0.720
## attitude 0.36066 0.05932 6.080 8.34e-09 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 5.315 on 162 degrees of freedom
## Multiple R-squared: 0.2018, Adjusted R-squared: 0.187
## F-statistic: 13.65 on 3 and 162 DF, p-value: 5.536e-08
Gender and age are not significantly associated to the points so remove them from the model. Fit the model again using only attitude as the explanatory variable.
lm_model_2 <- lm(points ~ attitude, data=learning2014)
summary(lm_model_2)
##
## Call:
## lm(formula = points ~ attitude, data = learning2014)
##
## Residuals:
## Min 1Q Median 3Q Max
## -16.9763 -3.2119 0.4339 4.1534 10.6645
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 11.63715 1.83035 6.358 1.95e-09 ***
## attitude 0.35255 0.05674 6.214 4.12e-09 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 5.32 on 164 degrees of freedom
## Multiple R-squared: 0.1906, Adjusted R-squared: 0.1856
## F-statistic: 38.61 on 1 and 164 DF, p-value: 4.119e-09
Explanatory variable attitude is statistically significantly associated to the outcome variable points with a p-value of 4.12e-09. The estimate of the regression coefficient is 0.35 meaning a one unit increase in attitude is on average associated to a 1.42 unit increase in points. The R² of the model is 0.19 which means that the variation of the explanatory variable attitude explains 19% of the variation of the outcome variable points.
Plot the residuals versus the fitted values
plot(lm_model_2, which=1)
Plot the normal QQ-plot:
plot(lm_model_2, which=2)
Plot the residuals versus leverage plot:
plot(lm_model_2, which=5)
The dataset has been constructed by joining two student alchol comsuption datasets which were downloaded from https://archive.ics.uci.edu/ml/datasets/Student+Performance. The two datasets contain the same variables and the students are partially overlapping. The data was joined by using following columns as surrogate identifiers for students:
The variables not used for joining have been combined by averaging for numeric columns and by taking the first answer for non-numeric columns. Two new variables have been define:
Read in the dataset.
alc <- read.csv("~/git/IODS-project/data/alc.csv")
Variables in the data set:
str(alc)
## 'data.frame': 382 obs. of 35 variables:
## $ school : Factor w/ 2 levels "GP","MS": 1 1 1 1 1 1 1 1 1 1 ...
## $ sex : Factor w/ 2 levels "F","M": 1 1 1 1 1 2 2 1 2 2 ...
## $ age : int 18 17 15 15 16 16 16 17 15 15 ...
## $ address : Factor w/ 2 levels "R","U": 2 2 2 2 2 2 2 2 2 2 ...
## $ famsize : Factor w/ 2 levels "GT3","LE3": 1 1 2 1 1 2 2 1 2 1 ...
## $ Pstatus : Factor w/ 2 levels "A","T": 1 2 2 2 2 2 2 1 1 2 ...
## $ Medu : int 4 1 1 4 3 4 2 4 3 3 ...
## $ Fedu : int 4 1 1 2 3 3 2 4 2 4 ...
## $ Mjob : Factor w/ 5 levels "at_home","health",..: 1 1 1 2 3 4 3 3 4 3 ...
## $ Fjob : Factor w/ 5 levels "at_home","health",..: 5 3 3 4 3 3 3 5 3 3 ...
## $ reason : Factor w/ 4 levels "course","home",..: 1 1 3 2 2 4 2 2 2 2 ...
## $ nursery : Factor w/ 2 levels "no","yes": 2 1 2 2 2 2 2 2 2 2 ...
## $ internet : Factor w/ 2 levels "no","yes": 1 2 2 2 1 2 2 1 2 2 ...
## $ guardian : Factor w/ 3 levels "father","mother",..: 2 1 2 2 1 2 2 2 2 2 ...
## $ traveltime: int 2 1 1 1 1 1 1 2 1 1 ...
## $ studytime : int 2 2 2 3 2 2 2 2 2 2 ...
## $ failures : int 0 0 2 0 0 0 0 0 0 0 ...
## $ schoolsup : Factor w/ 2 levels "no","yes": 2 1 2 1 1 1 1 2 1 1 ...
## $ famsup : Factor w/ 2 levels "no","yes": 1 2 1 2 2 2 1 2 2 2 ...
## $ paid : Factor w/ 2 levels "no","yes": 1 1 2 2 2 2 1 1 2 2 ...
## $ activities: Factor w/ 2 levels "no","yes": 1 1 1 2 1 2 1 1 1 2 ...
## $ higher : Factor w/ 2 levels "no","yes": 2 2 2 2 2 2 2 2 2 2 ...
## $ romantic : Factor w/ 2 levels "no","yes": 1 1 1 2 1 1 1 1 1 1 ...
## $ famrel : int 4 5 4 3 4 5 4 4 4 5 ...
## $ freetime : int 3 3 3 2 3 4 4 1 2 5 ...
## $ goout : int 4 3 2 2 2 2 4 4 2 1 ...
## $ Dalc : int 1 1 2 1 1 1 1 1 1 1 ...
## $ Walc : int 1 1 3 1 2 2 1 1 1 1 ...
## $ health : int 3 3 3 5 5 5 3 1 1 5 ...
## $ absences : int 5 3 8 1 2 8 0 4 0 0 ...
## $ G1 : int 2 7 10 14 8 14 12 8 16 13 ...
## $ G2 : int 8 8 10 14 12 14 12 9 17 14 ...
## $ G3 : int 8 8 11 14 12 14 12 10 18 14 ...
## $ alc_use : num 1 1 2.5 1 1.5 1.5 1 1 1 1 ...
## $ high_use : logi FALSE FALSE TRUE FALSE FALSE FALSE ...
I choose the following 4 variables for studying their relationship to high/low alcohol consumption:
I hypotize that high amount of “going out with friends” would be associated with a higer alcohol consumption since alcohol is often consumed in social situations. I think male gender could be associated to higher alcohol consumption just because males can tolerate more alcohol. I hypotize that higher “weekly study time” is associated to low alcohol compsuption: students with high study times are focused on school and don’t have as much time to drink alcohol. I also hypotize that individuals “with a romatic relationship” will consume less aclhol since they spend more time with their partners than friens and alchol is more often spend with friends.
Explore the variables of interest in regard to alchohol use
library(tidyr)
library(dplyr)
library(ggplot2)
# Explore the mean 'goout' and 'studytime' with respect to high/low alcohol use
alc %>% group_by(high_use) %>% summarise(count = n(), mean_goout=mean(goout),
mean_studytime=mean(studytime))
## # A tibble: 2 x 4
## high_use count mean_goout mean_studytime
## <lgl> <int> <dbl> <dbl>
## 1 FALSE 268 2.85 2.15
## 2 TRUE 114 3.72 1.77
# Explore high/low alcohol use stratified by 'sex'
alc %>% group_by(high_use, sex) %>% summarise(count = n())
## # A tibble: 4 x 3
## # Groups: high_use [?]
## high_use sex count
## <lgl> <fct> <int>
## 1 FALSE F 156
## 2 FALSE M 112
## 3 TRUE F 42
## 4 TRUE M 72
# Explore high/low alcohol use stratified by 'romantic'
alc %>% group_by(high_use, romantic) %>% summarise(count = n())
## # A tibble: 4 x 3
## # Groups: high_use [?]
## high_use romantic count
## <lgl> <fct> <int>
## 1 FALSE no 180
## 2 FALSE yes 88
## 3 TRUE no 81
## 4 TRUE yes 33
# Draw barplots of variables of interest
###########################################
g_goout <- ggplot(alc, aes(x = goout, fill=high_use)) +
geom_bar() + xlab("Going out with friends") +
ggtitle("Going out with friends from 1 (very low) to 5 (very high) by alcohol use")
g_studytime <- ggplot(alc, aes(x = studytime, fill=high_use)) +
geom_bar() + xlab("Weekly study time") +
ggtitle("Weekly study time [1 (<2 hours), 2 (2 to 5 hours), 3 (5 to 10 hours), or 4 (>10 hours)] by alchol use")
g_sex <- ggplot(alc, aes(x = sex, fill=high_use)) +
geom_bar() +
ggtitle("Sex by alcohol use")
g_romantic <- ggplot(alc, aes(x = romantic, fill=high_use)) +
geom_bar() +
ggtitle("With a romantic relationship (yes/no) by alcohol use")
# Arrange the plots into a grid
library("gridExtra")
grid.arrange(g_goout, g_studytime, g_sex, g_romantic, ncol=2, nrow=2)
Based on the plots my assumptions seem to be somewhat corret:
Fit a logistic regression model using high_use as the targe variable and goout, studytime, sex and romantic as the explanatory variables.
# Fit the logistic regression model
m <- glm(high_use ~ goout + studytime + sex + romantic, data = alc, family = "binomial")
Summary of the fitted logistics regression model shows that goout, studytime and sex are statistically significantly associated to alchol comsumption. High alcohol comsumption is associated to high goout and male gender and low alcohol comsumption is associated to high studytime.
# Summary of the model
summary(m)
##
## Call:
## glm(formula = high_use ~ goout + studytime + sex + romantic,
## family = "binomial", data = alc)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -1.7365 -0.8114 -0.5009 0.9081 2.6642
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) -2.6988 0.5712 -4.725 2.30e-06 ***
## goout 0.7536 0.1187 6.350 2.15e-10 ***
## studytime -0.4774 0.1683 -2.837 0.00456 **
## sexM 0.6657 0.2585 2.576 0.01000 *
## romanticyes -0.1424 0.2699 -0.528 0.59767
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 465.68 on 381 degrees of freedom
## Residual deviance: 393.67 on 377 degrees of freedom
## AIC: 403.67
##
## Number of Fisher Scoring iterations: 4
Coefficients of the model as odds ratios and their confidence intervals:
# Calculate the odds ratios and confidence intervals of the coefficients
or <- coef(m) %>% exp
ci <- confint(m) %>% exp
# Print out the odds ratios and confidence intervals
cbind(or, ci)
## or 2.5 % 97.5 %
## (Intercept) 0.06728696 0.02129867 0.2010636
## goout 2.12456419 1.69404697 2.7003422
## studytime 0.62040325 0.44145946 0.8558631
## sexM 1.94589655 1.17538595 3.2443631
## romanticyes 0.86724548 0.50714091 1.4648961
From the odds ratios we can see that one unit increase in goout is associated with 2.1 times higher likelihood of high alchohol comsumption and one unit increase in studytime is associated with 0.6 times lower likelihood of high alcohol comsumption. Male gender is associated with 1.9 higer likelihood of high alcohol comsumption comapred to female gender. Being in a romatic relationship is not significantly associated to high/low alcohol comsuption in this model since the confidence intervals include 1.
My previously stated hypothesis seem to be verified by this model except that romantic is not associated to high or low alcohol comsumption in this model.
Fit a logistic model with the explanatory variables that were statistically significantly associated to high or low alchohol consumption:
# Fit the logistic regression model
m <- glm(high_use ~ goout + studytime + sex, data = alc, family = "binomial")
Prediction performance of the model
# Calcualte the predicted probabilities of high alcohol comspumtion
probability <- predict(m, type="response")
alc <- mutate(alc, probability=probability)
# Predict the high alcohol use with the probabilities
alc <- mutate(alc, prediction=probability > 0.5)
# Cross-tabulate the actual class and the predicted class
table(high_use = alc$high_use, prediction = alc$prediction)
## prediction
## high_use FALSE TRUE
## FALSE 250 18
## TRUE 76 38
The model seems to be quite good at predicting low alcohol use but performes less well in predicting high alcohol use.
Visualize the actual class, the predicted probabilities and the predicted class.
# Initialize a plot of 'high_use' versus 'probability' in 'alc'
g <- ggplot(alc, aes(x = probability, y = high_use, col=prediction))
# define the geom as points and draw the plot
g + geom_point()
Calculate the total proportion of mis-classified individuals using the regression model and with a simple guessing strategy where everyone is classified to be in the most prevalent class low use of alcohol.
# Define a loss function (mean prediction error)
loss_func <- function(class, prob) {
n_wrong <- abs(class - prob) > 0.5
mean(n_wrong)
}
# Call loss_func to compute the average number of wrong predictions in the (training) data
loss_func(class = alc$high_use, prob = alc$probability)
## [1] 0.2460733
# Compare the results to guessing that everybody belongs to the class low use of alcohol
loss_func(class = alc$high_use, prob = 0)
## [1] 0.2984293
Using the regression model 24.6% of the individuals are mis-classified compared to 29.8 % of mis-classified individuals when simply guessing everybody belongs to the low use of alcohol class. The model seems to provide some improvement to the simple guess of the most prevalent class.
Performe 10-fold cross-validation of the model to estiamte the performance of the model on unseen data. The performance of the model is measured with proportion of mis-classified individuals. The mean prediction error in the test set:
library(boot)
# 10-fold cross-validation
cv <- cv.glm(data = alc, cost = loss_func, glmfit = m, K = 10)
# Mean prediction error
cv$delta[1]
## [1] 0.2460733
The mean prediction error in the test set is 0.25 which is better than the performance of the model introduced in the DataCamp exercises which had a mean prediction error of 0.26 in the test set.
Construct models with different number of predictors and calculate the test set and training set prediction errors.
# All the possible predictors
predictors <- c('school', 'sex', 'age', 'address', 'famsize', 'Pstatus', 'Medu',
'Fedu', 'Mjob', 'Fjob', 'reason', 'nursery', 'internet', 'guardian',
'traveltime', 'studytime', 'failures', 'schoolsup', 'famsup', 'paid',
'activities', 'higher', 'romantic', 'famrel', 'freetime', 'goout',
'health', 'absences', 'G1', 'G2', 'G3')
# Fit several models and record the test and traingin errors
# 1) Use all of the predictors.
# 2) Drop one predictor and fit a new model.
# 3) Continue until only one predictor is left in the model.
# Fit the models and calculate the erros
test_error <- numeric(length(predictors))
training_error <- numeric(length(predictors))
for(i in length(predictors):1) {
model_formula <- paste0("high_use ~ ", paste(predictors[1:i], collapse = " + "))
glmfit <- glm(model_formula, data = alc, family = "binomial")
# 10-fold cross-validation
cv <- cv.glm(data = alc, cost = loss_func, glmfit = m, K = 10)
# Mean prediction error
test_error[i] <- cv$delta[1]
# Training error
training_error[i] <-
loss_func(alc$high_use, predict(glmfit,type="response"))
}
# Construct a table of prediction errors for plotting
data_error <- rbind(data.frame(n_predictors=1:length(predictors),
prediction_error=test_error,
type = "test error"),
data.frame(n_predictors=1:length(predictors),
prediction_error=training_error,
type = "training error"))
# Plot the test and training errors vs. number of predictors in the model
g <- ggplot(data_error, aes(x = n_predictors, y = prediction_error, col=type))
# define the geom as points and draw the plot
g + geom_point()
# Load libraries
library(corrplot)
library(dplyr)
Load in the Boston dataset from the MASS package.
library(MASS)
data("Boston")
str(Boston)
## 'data.frame': 506 obs. of 14 variables:
## $ crim : num 0.00632 0.02731 0.02729 0.03237 0.06905 ...
## $ zn : num 18 0 0 0 0 0 12.5 12.5 12.5 12.5 ...
## $ indus : num 2.31 7.07 7.07 2.18 2.18 2.18 7.87 7.87 7.87 7.87 ...
## $ chas : int 0 0 0 0 0 0 0 0 0 0 ...
## $ nox : num 0.538 0.469 0.469 0.458 0.458 0.458 0.524 0.524 0.524 0.524 ...
## $ rm : num 6.58 6.42 7.18 7 7.15 ...
## $ age : num 65.2 78.9 61.1 45.8 54.2 58.7 66.6 96.1 100 85.9 ...
## $ dis : num 4.09 4.97 4.97 6.06 6.06 ...
## $ rad : int 1 2 2 3 3 3 5 5 5 5 ...
## $ tax : num 296 242 242 222 222 222 311 311 311 311 ...
## $ ptratio: num 15.3 17.8 17.8 18.7 18.7 18.7 15.2 15.2 15.2 15.2 ...
## $ black : num 397 397 393 395 397 ...
## $ lstat : num 4.98 9.14 4.03 2.94 5.33 ...
## $ medv : num 24 21.6 34.7 33.4 36.2 28.7 22.9 27.1 16.5 18.9 ...
The dataset has 14 variables and 506 observations. The following variables are present:
More details of the dataset can be found here https://stat.ethz.ch/R-manual/R-devel/library/MASS/html/Boston.html.
Summary of the variabes in the dataset:
summary(Boston)
## crim zn indus chas
## Min. : 0.00632 Min. : 0.00 Min. : 0.46 Min. :0.00000
## 1st Qu.: 0.08204 1st Qu.: 0.00 1st Qu.: 5.19 1st Qu.:0.00000
## Median : 0.25651 Median : 0.00 Median : 9.69 Median :0.00000
## Mean : 3.61352 Mean : 11.36 Mean :11.14 Mean :0.06917
## 3rd Qu.: 3.67708 3rd Qu.: 12.50 3rd Qu.:18.10 3rd Qu.:0.00000
## Max. :88.97620 Max. :100.00 Max. :27.74 Max. :1.00000
## nox rm age dis
## Min. :0.3850 Min. :3.561 Min. : 2.90 Min. : 1.130
## 1st Qu.:0.4490 1st Qu.:5.886 1st Qu.: 45.02 1st Qu.: 2.100
## Median :0.5380 Median :6.208 Median : 77.50 Median : 3.207
## Mean :0.5547 Mean :6.285 Mean : 68.57 Mean : 3.795
## 3rd Qu.:0.6240 3rd Qu.:6.623 3rd Qu.: 94.08 3rd Qu.: 5.188
## Max. :0.8710 Max. :8.780 Max. :100.00 Max. :12.127
## rad tax ptratio black
## Min. : 1.000 Min. :187.0 Min. :12.60 Min. : 0.32
## 1st Qu.: 4.000 1st Qu.:279.0 1st Qu.:17.40 1st Qu.:375.38
## Median : 5.000 Median :330.0 Median :19.05 Median :391.44
## Mean : 9.549 Mean :408.2 Mean :18.46 Mean :356.67
## 3rd Qu.:24.000 3rd Qu.:666.0 3rd Qu.:20.20 3rd Qu.:396.23
## Max. :24.000 Max. :711.0 Max. :22.00 Max. :396.90
## lstat medv
## Min. : 1.73 Min. : 5.00
## 1st Qu.: 6.95 1st Qu.:17.02
## Median :11.36 Median :21.20
## Mean :12.65 Mean :22.53
## 3rd Qu.:16.95 3rd Qu.:25.00
## Max. :37.97 Max. :50.00
Explore the distribution of the variables by plotting:
library(GGally)
library(ggplot2)
# Define the plot.
p <- ggpairs(Boston, mapping = aes(alpha=0.3),
lower = list(combo = wrap("facethist", bins = 20)))
# Draw the plot
p
Correlation of the variables:
cor(Boston) %>% corrplot(method="circle", type="upper", cl.pos="b", tl.pos="d")
Scale the dataset so that the mean of each variable is zero and standard deviation is one:
\[x_{scaled}=\frac{x - \mu_{x}}{\sigma_{x}}\],
where \(\mu_{x}\) is the mean of x and \(\sigma_{x}\) the standard deviation of x.
boston_scaled <- scale(Boston) %>% as.data.frame()
summary(boston_scaled)
## crim zn indus
## Min. :-0.419367 Min. :-0.48724 Min. :-1.5563
## 1st Qu.:-0.410563 1st Qu.:-0.48724 1st Qu.:-0.8668
## Median :-0.390280 Median :-0.48724 Median :-0.2109
## Mean : 0.000000 Mean : 0.00000 Mean : 0.0000
## 3rd Qu.: 0.007389 3rd Qu.: 0.04872 3rd Qu.: 1.0150
## Max. : 9.924110 Max. : 3.80047 Max. : 2.4202
## chas nox rm age
## Min. :-0.2723 Min. :-1.4644 Min. :-3.8764 Min. :-2.3331
## 1st Qu.:-0.2723 1st Qu.:-0.9121 1st Qu.:-0.5681 1st Qu.:-0.8366
## Median :-0.2723 Median :-0.1441 Median :-0.1084 Median : 0.3171
## Mean : 0.0000 Mean : 0.0000 Mean : 0.0000 Mean : 0.0000
## 3rd Qu.:-0.2723 3rd Qu.: 0.5981 3rd Qu.: 0.4823 3rd Qu.: 0.9059
## Max. : 3.6648 Max. : 2.7296 Max. : 3.5515 Max. : 1.1164
## dis rad tax ptratio
## Min. :-1.2658 Min. :-0.9819 Min. :-1.3127 Min. :-2.7047
## 1st Qu.:-0.8049 1st Qu.:-0.6373 1st Qu.:-0.7668 1st Qu.:-0.4876
## Median :-0.2790 Median :-0.5225 Median :-0.4642 Median : 0.2746
## Mean : 0.0000 Mean : 0.0000 Mean : 0.0000 Mean : 0.0000
## 3rd Qu.: 0.6617 3rd Qu.: 1.6596 3rd Qu.: 1.5294 3rd Qu.: 0.8058
## Max. : 3.9566 Max. : 1.6596 Max. : 1.7964 Max. : 1.6372
## black lstat medv
## Min. :-3.9033 Min. :-1.5296 Min. :-1.9063
## 1st Qu.: 0.2049 1st Qu.:-0.7986 1st Qu.:-0.5989
## Median : 0.3808 Median :-0.1811 Median :-0.1449
## Mean : 0.0000 Mean : 0.0000 Mean : 0.0000
## 3rd Qu.: 0.4332 3rd Qu.: 0.6024 3rd Qu.: 0.2683
## Max. : 0.4406 Max. : 3.5453 Max. : 2.9865
From the summary we can see that the mean of the scaled variables is zero.
Create a factor variable críme from the crim (per capita crime rate by town) by dividing the crim variable by quartiles to ‘low’, ‘med_low’, ‘med_high’ and ‘high’ categories:
# Create a quantile vector of crim and print it
bins <- quantile(boston_scaled$crim)
# Create a categorical variable 'crime'
crime <- cut(boston_scaled$crim, breaks = bins, include.lowest = TRUE,
label=c("low", "med_low", "med_high", "high"))
# Remove original crim from the dataset
boston_scaled <- dplyr::select(boston_scaled, -crim)
# Add the new categorical value to scaled data
boston_scaled <- data.frame(boston_scaled, crime)
Divide the dataset to training and test sets so that 80% belongs to the training set and 20% belongs to the test set.
# Set seed so the results are reproducible
set.seed(1234)
# Take randomly 80% of the observations to the training set
train.idx <- sample(nrow(boston_scaled), size = 0.8 * nrow(boston_scaled))
train <- boston_scaled[train.idx,]
# Take the remaining 20% to the test set
test <- boston_scaled[-train.idx,]
Fit the linear discriminant analysis (LDA) on the training set using the categorical crime rate as the target variable and all the other variables in the dataset as predictor variables.
# linear discriminant analysis
lda.fit <- lda(crime ~ ., data = train)
The LDA biplot:
# the function for lda biplot arrows
lda.arrows <- function(x, myscale = 1, arrow_heads = 0.1, color = "red", tex = 0.75, choices = c(1,2)){
heads <- coef(x)
arrows(x0 = 0, y0 = 0,
x1 = myscale * heads[,choices[1]],
y1 = myscale * heads[,choices[2]], col=color, length = arrow_heads)
text(myscale * heads[,choices], labels = row.names(heads),
cex = tex, col=color, pos=3)
}
# target classes as numeric
classes <- as.numeric(train$crime)
# plot the lda results
plot(lda.fit, dimen = 2, col=classes, pch=classes)
lda.arrows(lda.fit, myscale = 2)
Use the fitted LDA model to predict the categorical crime rate in the test set. Cross tabulate the observed classes and the predicted classes in the test set:
# Save the correct classes from test data
correct_classes <- test$crime
# Remove the crime variable from test data
test <- dplyr::select(test, -crime)
# predict classes with test data
lda.pred <- predict(lda.fit, newdata = test)
# cross tabulate the results
table(correct = correct_classes, predicted = lda.pred$class)
## predicted
## correct low med_low med_high high
## low 11 8 0 0
## med_low 7 19 2 0
## med_high 1 6 18 1
## high 0 0 0 29
Model seems to perform perfectly at predicting the ‘high’ class and also predicts the other classes reasonably well. The prediction accuracy is worst for the ‘low’ class. The model mis-classifies a big proportion of the ‘low’ observations as ‘med_low’.
Reload the Boston dataset and standardize it as before. Calculate the Euclidean distance between the observations:
# Load the Boston dataset
data("Boston")
# Scale the dataset
boston_scaled <- scale(Boston) %>% as.data.frame()
# Calculate the Euclidean distance between the pairs of observations
dist_eu <- dist(boston_scaled)
summary(dist_eu)
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 0.1343 3.4620 4.8240 4.9110 6.1860 14.4000
Run the k-means algorithm with 3 clusters and visualize the results:
# Set seed to get reproducible results
set.seed(123)
# k-means clustering
km <-kmeans(boston_scaled, centers = 3)
# plot the Boston dataset with clusters
pairs(boston_scaled, col = km$cluster)
Calculate te total of within cluster sum of squares (TWCSS) when the number of cluster changes from 1 to 10.
# Set seed to get reproducible results
set.seed(123)
# Determine the number of clusters
k_max <- 10
# Calculate the total within sum of squares
twcss <- sapply(1:k_max, function(k){kmeans(boston_scaled, k)$tot.withinss})
# visualize the results
qplot(x = 1:k_max, y = twcss, geom = 'line')
The optimal number of clusters is when the total WCSS drops radically so based on the graph 2 seems to be the optimal number of clusters. Perform k-means with 2 clusters and visualize the results.
# Set seed to get reproducible results
set.seed(123)
# k-means clustering
km <-kmeans(boston_scaled, centers = 2)
# plot the Boston dataset with clusters
pairs(boston_scaled, col = km$cluster)
Perform k-means clustering with 3 clusters on the scaled Boston datset. Use the cluster asigments as the target variable for LDA analysis.
# Set seed to get reproducible results
set.seed(123)
# k-means clustering
km <-kmeans(boston_scaled, centers = 3)
# Add the cluster assingment to the dataset
boston_scaled$kmeans_cluster <- km$cluster
# linear discriminant analysis
lda.fit <- lda(kmeans_cluster ~ ., data = boston_scaled)
The LDA biplot:
# plot the lda results
plot(lda.fit, dimen = 2, col=boston_scaled$kmeans_cluster,
pch=boston_scaled$kmeans_cluster)
lda.arrows(lda.fit, myscale = 2)
Based on the biplot the most influencal linear separators are: